What is the extreme f(x,y)=x^3+y^3-6xy ?

The extreme f(x,y)=x^3+y^3-6xy is Saddle(0,0),Minimum(2,2)

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extreme f(x,y)=x^3+y^3-6xy

extreme points

f (x, y) = x^{3} + y^{3} - 6 x y

Saddle (0, 0), Minimum (2, 2)

Solution steps

Find the critical points:

(0, 0), (2, 2)

\frac{\partial ^{2} f}{\partial x ^{2}} = 6 y

D (x, y) = 36 x y - 36

Check the sign of at each critical point

Check critical point

(0, 0) :

Saddle

Check critical point

(2, 2) :

Minimum

Saddle (0, 0), Minimum (2, 2)

f (x) = 2 x + cos (2 x)

f (x) = \frac{1}{3} x^{3} - x^{2} - 3 x + 1

f (x) = \frac{1}{3} x^{3} - x^{2} - 3 x + 4

f (x) = x^{3} - 3 x + 3 x y^{2}

f (x, y) = 2 x^{2} + y^{2} - 3 x

What is the extreme f(x,y)=x^3+y^3-6xy ?
The extreme f(x,y)=x^3+y^3-6xy is Saddle(0,0),Minimum(2,2)